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Energy

I did gpe=mgh=ke

so for first one:

mgh=(250x9.81x57)=139792.5

Sqrt(((139792.5)x2)/(250))=v=33.44

i got that for al of them, is this correct?
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Reply 1
Avatar for Mutleybm1996
Mutleybm1996
OP
19
It doesn't seem correct but the maths is sound
Reply 2
Avatar for Hummi_C
Hummi_C
14
If you used the same method, then yes you are correct
Reply 3
Avatar for Mutleybm1996
Mutleybm1996
OP
19
Original post by Hummi_C
If you used the same method, then yes you are correct


Will it be the same for the next few as well? Surely the lighter people have a lesser velocity?
Reply 4
Avatar for alow
alow
19
Original post by benwalters1996
Will it be the same for the next few as well? Surely the lighter people have a lesser velocity?


Why? All falling objects accelerate at the same rate, assuming no drag forces.
(edited 10 years ago)
Reply 5
Avatar for Mutleybm1996
Mutleybm1996
OP
19
Original post by alow
Why? All objects accelerate at the same rate, assuming no drag forces.


How would i work out efficiency?
Reply 6
Avatar for alow
alow
19
Original post by benwalters1996
How would i work out efficiency?


100×(total energy beforetotal energy after)100 \times \left( \dfrac{\text{total\ energy\ before}}{\text{total\ energy\ after}} \right)
Reply 7
Avatar for Hummi_C
Hummi_C
14
Original post by benwalters1996
Will it be the same for the next few as well? Surely the lighter people have a lesser velocity?


It will be the same, because:
GPE=EKGPE=E_K
mgh=12mv2mgh=\frac{1}{2}mv^2
Therefore
2mgh=mv22mgh=mv^2
2mghm=mv2m[br][br][br]2gh=v2\frac{2mgh}{m}=\frac{mv^2}{m}[br][br][br]2gh=v^2
So
v=2ghv=\sqrt{2gh}
Since g and h are constants for all objects, 2gh\sqrt{2gh} can be written as a constant, cc
Therefore
v=cv=c for all three objects.
Understand? :smile:
Reply 8
Avatar for Hummi_C
Hummi_C
14
Original post by benwalters1996
How would i work out efficiency?



Original post by alow
100×(total energy beforetotal energy after)100 \times \left( \dfrac{\text{total\ energy\ before}}{\text{total\ energy\ after}} \right)


It's actually
100×(total energy aftertotal energy before)100 \times \left( \dfrac{\text{total\ energy\ after}}{\text{total\ energy\ before}} \right)
Reply 9
Avatar for lerjj
lerjj
13
Given each of their masses, you can work out GPE at the point at which they fall from and for their rebound heights. However, since this is going to give you something like E=100%*mgh1/mgh2, most of that cancels so efficiency is just ratio of rebound height to origional height.

Edit: You can also find v in the first place by using SUVAT v2=u2+2as, where u=0, a=g and s=h. No kinetic energy forumla needed at all!
(edited 10 years ago)

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